Let \(g\mapsto \Lambda _g\)  be the canonical isometric isomorphism identification of \(L^q(G)\)  as \(L^p(G)^*\) , where \(\Lambda _g:L^p(G)\rightarrow \mathbb {C}\)  is given by \(\Lambda _g(f):=\langle f,g\rangle \)  for every \(f\in L^p(G)\) .
Invoking Theorem REF , we conclude that the dual space
\(L^p_\xi (G,H)^*\)  is isometrically isomorphic to \(\mathcal {N}_\xi ^p(G,H)^\perp \) , where
\(\mathcal {N}_\xi ^p(G,H)^\perp =\left\lbrace g\in L^q(G):\Lambda _g(f)=0,\ \ {\rm for\ all}\ f\in \mathcal {N}_\xi ^p(G,H)\right\rbrace .\) 
We then claim that \(\mathcal {N}_\xi ^p(G,H)^\perp =L^q_\xi (G,H)\) . To show this, let \(g\in \mathcal {N}_\xi ^p(G,H)^\perp \)  be given. Then, \(g\in L^q(G)\)  and
\(\Lambda _g(f)=0\)  for all \(f\in \mathcal {N}_\xi ^p(G,H)\) . Suppose that \(h\in H\)  is arbitrary. Then, using (REF ) and Theorem REF , for every \(f\in \mathcal {C}_c(G)\)  we have \(R_{h^{-1}}f-\overline{\xi (h)}f\in \mathcal {N}_\xi ^p(G,H)\) . This implies that \(\Lambda _g(R_{h^{-1}}f)=\overline{\xi (h)}\Lambda _g(f)\) . Therefore, using compactness of \(H\) , we get
\(\Lambda _{R_hg}(f)&=\int _Gf(x)\overline{g(xh)}\mathrm {d}\lambda _G(x)\\&=\int _Gf(xh^{-1})\overline{g(x)}\mathrm {d}\lambda _G(xh^{-1})\\&=\Delta _G(h^{-1})\int _Gf(xh^{-1})\overline{g(x)}\mathrm {d}\lambda _G(x)\\&=\Delta _H(h^{-1})\int _Gf(xh^{-1})\overline{g(x)}\mathrm {d}\lambda _G(x)\\&=\int _Gf(xh^{-1})\overline{g(x)}\mathrm {d}\lambda _G(x)\\&=\Lambda _g(R_{h^{-1}}f)=\overline{\xi (h)}\Lambda _g(f)=\Lambda _{\xi (h)g}(f).\) 

Since \(f\in \mathcal {C}_c(G)\)  was arbitrary, we get \(R_hg=\xi (h)g\)  in \(L^q(G)\) . Since \(h\in H\)  was arbitrary, we conclude that \(g\in L^q_\xi (G,H)\) . Conversely, suppose that \(g\in L^q_\xi (G,H)\) . We shall show that \(\Lambda _g(f)=0\)  for every
\(f\in \mathcal {N}_\xi ^p(G,H)\) . According to Theorem 2.49 of [1]}, let \(\lambda _{G/H}\)  be the \(G\) -invariant Radon measure on the left coset space \(G/H\)  normalized with respect to the Weil's formula (2.50) of [1]}.
Then, for every \(f\in \mathcal {N}_\xi ^p(G,H)\) , we have
\(\Lambda _g(f)&=\int _G f(x)\overline{g(x)}\mathrm {d}\lambda _G(x)\\&=\int _{G/H}\left(\int _Hf(xh)\overline{g(xh)}\mathrm {d}\lambda _H(h)\right)\mathrm {d}\lambda _{G/H}(xH)\\&=\int _{G/H}\left(\int _Hf(xh)\overline{\xi (h)}\mathrm {d}\lambda _H(h)\right)\overline{g(x})\mathrm {d}\lambda _{G/H}(xH)=\int _{G/H}T_\xi (f)(x)\overline{g(x})\mathrm {d}\lambda _{G/H}(xH)=0,\) 

which implies that \(g\in \mathcal {N}_\xi ^p(G,H)^\perp \) .
